SOLUTION
We are told that the rectangle is 5 times as long as it is wide. This means that the length of the rectangle is 5 times its width.
Let the length of the rectangle be L and let the width be w, then, it means that
[tex]\begin{gathered} L=5\times w \\ L=5w \end{gathered}[/tex]
Also we were told the area is 20 square-feet. But area of a rectangle is calculated as
[tex]\begin{gathered} Area=length\times width \\ A=L\times w \\ We\text{ have L = 5w, so } \\ A=5w\times w \\ A=5w^2,\text{ but the area is 20 ft}^2,\text{ so } \\ 20=5w^2 \end{gathered}[/tex]
Solving for w in the equation we just made, we have
[tex]\begin{gathered} 20=5w^2 \\ dividing\text{ by 5} \\ \frac{20}{5}=\frac{5w^2}{5} \\ 4=w^2 \\ w^2=4 \\ take\text{ the square to 4, it becomes square root, we have } \\ w=\sqrt{4} \\ w=2\text{ ft} \end{gathered}[/tex]
So the width is 2 ft, the length becomes
[tex]\begin{gathered} L=5w \\ L=5\times2 \\ L=10\text{ ft } \end{gathered}[/tex]
Perimeter of a rectangle is calculated using the formula
[tex]\begin{gathered} Perimeter=2(L+w) \\ P=2(L+w) \\ P=2(10+2) \\ P=2(12) \\ P=2\times12 \\ P=24\text{ ft } \end{gathered}[/tex]
Hence the perimeter is 24 ft
